# Convergent and Divergent Series Examples

Lets learn first what is the **convergent** and **divergent** series.

## What is convergent series and divergent series ?

A series which have finite sum is called convergent series.Otherwise is called divergent series.

$ \lim _{n \rightarrow \infty} S_{n}=S $

- If the partial sums Sn of an infinite series tend to a limit S, the series is called
convergent.

Otherwise it is called divergent. - The limiting value S is called the sum of the series

Lets look at some examples of convergent and divergence series examples.

Let us consider two series

## Convergent series example

The sum of this series is finite.

$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots $

## Divergent series example

The sum of this series is not finite.

$ 1+2+4+8+16+\cdots $

## Tests for convergence of series

It can be difficult to know if a series is convergent or divergent.

But here is some methods that can be used to determine if the series is convergent or divergent

- Preliminary test
- Comparison test
- Ratio test

### Preliminary Test

A necessary but not sufficient condition for a series of real positive terms $ \sum u_{n} $to be convergent is that the term u_{n}tends to zero as n tends to infinity ie.

$ \lim _{n \rightarrow \infty} u_{n}=0 $

If this condition does not satisfy then series must diverge.But if this condition get satisfied then series can be divergent or convergent because this is not a sufficient condition for convergence.

### Comparison test

In comparison test we compare our series with a series whose convergence is already known to us.

Let us consider two series $ \sum u_{n} $ and $ \sum u_{n} $ and suppose that we know the latter to be convergent.

Then, if each term u_{n} in the first series is equal to less than the each term in second series v_{n}for all n greaterthan some fixed number N that will vary from series to series,

then the original series $ \sum u_{n} $ is also **convergent.**

In other words if series $\sum v_{n} $ is convergent and,

$ u_{n} \leq v_{n} \quad $ for $ n>N $ then, $ \sum u_{n} $ converges.

However,if $ \sum v_{n} $ diverges and $ u_{n} \geq v_{n} $ for all $ n $ greater than some fixed number then $ \sum u_{n} $ diverges.

### Example #1

#### Determine whether following series is divergent or convergent.

$ \sum_{n=1}^{\infty} \frac{1}{n !+1}$$=\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{25}+\cdots $

We compare this series with the series of $ \sum_{n=0}^{\infty} \frac{1}{n !} $

$ \sum_{n=0}^{\infty} \frac{1}{n !}=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots=2+\frac{1}{2 !}+\frac{1}{3 !}+\cdots $

Here this series is perfectly equals to expension of $e^{x}$ when x=1 i.e

$e^{1}=e=$$ 1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots $

Now we know that the series of e is convergent.Because it has a finite sum.

Since each term in the first series is less than the corresponding term in second series.

So first series is also **convergent**.

### Ratio test

The ratio test determine whether a series converges by comparing the relative magnitudes of successive terms.

If we consider a series $\sum u_{n}$ then ratio of successive terms is,

$ \rho=\lim _{n \rightarrow \infty}\left(\frac{u_{n+1}}{u_{n}}\right) $

**Convergent** : if $ \rho \lt 1 $

**Divergent** : if $ \rho \gt 1 $

**Undetermined** : if $\rho = 1$

### Example #2

#### Determine whether following series is divergent or convergent.

$ \sum_{n=0}^{\infty} \frac{1}{n !}=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots$$=2+\frac{1}{2 !}+\frac{1}{3 !}+\cdots $

By using ratio test we have,

$ \rho=\lim _{n \rightarrow \infty}\left[\frac{n !}{(n+1) !}\right]$$=\lim _{n \rightarrow \infty}\left(\frac{1}{n+1}\right)=0 $

Since $\rho \lt 1 $ so series **converges**.